What is the largest integer $n$ for which $\binom{8}{3} + \binom{8}{4} = \binom{9}{n}$?
Explanation: By Pascal's Identity, we have $\binom{8}{3} + \binom{8}{4} = \binom{9}{4}$.  However, we also have $\binom{9}{4} = \binom{9}{9-4} = \binom{9}{5}$.  There are no other values of $n$ such that $\binom{9}{4} = \binom{9}{n}$, so the largest possible value of $n$ is $\boxed{5}$.